codeforces535C:Tavas and Karafs（二分）

Karafs is some kind of vegetable in shape of an 1 × h rectangle. Tavaspolis people love Karafs and they use Karafs in almost any kind of food. Tavas, himself, is crazy about Karafs.

Each Karafs has a positive integer height. Tavas has an infinite 1-based sequence of Karafses. The height of the i-th Karafs is si = A + (i - 1) × B.

For a given m, let's define an m-bite operation as decreasing the height of at most m distinct not eaten Karafses by 1. Karafs is considered as eaten when its height becomes zero.

Now SaDDas asks you n queries. In each query he gives you numbers l, t and m and you should find the largest number r such thatl ≤ r and sequence sl, sl + 1, ..., sr can be eaten by performing m-bite no more than t times or print -1 if there is no such number r.

Input

The first line of input contains three integers A, B and n (1 ≤ A, B ≤ 106, 1 ≤ n ≤ 105).

Next n lines contain information about queries. i-th line contains integers l, t, m (1 ≤ l, t, m ≤ 106) for i-th query.

Output

For each query, print its answer in a single line.

Sample Input

Input

2 1 4
1 5 3
3 3 10
7 10 2
6 4 8

Output

4
-1
8
-1

Input

1 5 2
1 5 10
2 7 4

Output

1
2

首先。。你要看懂英文题目是什么意思（CF里的题像我这种渣渣就是要看半天题orz。。。）

题目给出a,b,n，是一个由a开头，b为公差，长度无限的等差数列。。（数学题？）再输入l,t,m，取这个数列从l开始，m个数，每取一次这个数列中所有的数字-1，当首尾变成0的时候，可以向后移，问最后t次之后，最长的0序列的右边界是多少？

看着貌似比较难的样子，。。。别急。。慢慢一点一点分析。。。

首先可以确定，对于这个最终的序列而言，假设右边界为r，那么我们可以挖掘出这样两个条件。。。

max(h1,h2,...hr)<=t   

 和

h1+h2+...hr<=t*m
对于这两个条件，我们进行二分。。。。

（还是应了那句话，不是所有的二分都是裸二分。。。只出一个二分模板题是不可能的、。。。。）

上代码。。

#include<stdio.h>#include<ctype.h>#include<string.h>#include<stdlib.h>#include<math.h>#define MAXN 1000+10long long  a,b,n;  long long l,t,m;    long long cal(long long x)  {      return a+(x-1)*b;  }    long long get_sum(long long r)  {      return (cal(r)+cal(l))*(r-l+1)/2;  }    int main()  {      long long i,j,k,maxn;      while(~scanf("%I64d%I64d%I64d",&a,&b,&n))      {          while(n--)          {              scanf("%I64d%I64d%I64d",&l,&t,&m);              if(cal(l)>t)              {                  printf("-1\n");                  continue;              }              long long ll  = l,lr = (t-a)/b+1,mid;              while(ll<=lr)              {                  long long mid = (ll+lr)/2;                  if(get_sum(mid)<=t*m) ll = mid+1;                  else lr = mid-1;              }              printf("%d\n",ll-1);          }      }        return 0;  }